The value of 'a' for which one root of by equation (a^2-5a+3)x^2 +(3a-1)x +2=0 is twice as large as the other is

Question

This question belongs to the chapter QUADRATIC EXPRESSIONS

Answer 1

In a quadratic equation we can write the factors as (x-A)(x-B)=0 where A and B are the roots. In this case B=2A.

When this is expanded we get x^2-(A+B)x+AB=0=x^2-3Ax+2A^2=0.

Divide the given equation by a^2-5a+3: x^2+(3a-1)x/(a^2-5a+3)+2/(a^2-5a+3)=0.

Equating terms we get: -3A=(3a-1)/(a^2-5a+3) and 2A^2=2/(a^2-5a+3), A^2=1/(a^2-5a+3)
 
So, 9A^2=(3a-1)^2/(a^2-5a+3)^2 and 9/(a^2-5a+3)=(3a-1)^2/(a^2-5a+3)^2.

9(a^2-5a+3)=(3a-1)^2=9a^2-6a+1.

9a^2-45a+27=9a^2-6a+1; -45a+27=-6a+1; 26=39a so a=26/39=2/3.

(The original equation becomes: x^2/9+x+2=0 or x^2+9x+18=0=(x+6)(x+3) whose roots are -3 and -6, one root being twice the other. So a=2/3 is proven.)

Answer 2

Given (a^2-5a+3)x^2 +(3a-1)x +2=0

Value one root = A and

Second root B = 2A

Factores of equation
(x-A)(x-B)=0
=>x^2-(A+B)x+AB=0
=>x^2-3Ax+2A^2=0
=> x^2+(3a-1)x/(a^2-5a+3)+2/(a^2-5a+3)=0
=> -3A=(3a-1)/(a^2-5a+3)
=> 2A^2=2/(a^2-5a+3),
=> A^2=1/(a^2-5a+3)
9(a^2-5a+3)=(3a-1)^2=9a^2-6a+1
a=26/39=2/3

So the root of the equation are
(x+6)(x+3)=0
x=-3 and -6

Quadratic Equations